Plane projective curves and their components

Fix an algebraically closed field $k$ An algebraically closed field: every nonconstant polynomial has a root in the field and ${\mathbf{P}}^2={\mathbf{P}}_k^2$ projective space points. A (plane) projective curve is a closed subset $C=V(F)\subseteq {\mathbf{P}}^2$ given by a nonconstant square-free homogeneous form $F\in k[x_0,x_1,x_2]$ homogeneous polynomial and homogeneous ideal, projective algebraic set, projective zariski topology. Its degree is the total degree $\mathrm{deg}F$ Monomials, coefficients, degree in each variable and total degree in $F[x_1,\dots ,x_n]$, and its irreducible components are the closed sets $V(h)$ for the distinct irreducible factors $h$ of $F$, each taken once. Thus every component is reduced and occurs with multiplicity one, $C$ may be reducible, and $C$ is nonempty. A closed set $V(F)$ with $F$ nonconstant and square-free is written $V(F)$, and we call $F$ a defining form of $C$; under the Axiom of Choice, it is determined by $C$ up to a nonzero scalar: the square-free principal ideal $(F)$ is radical, and the homogeneous radical-ideal correspondence recovers it from $V(F)$ projective irreducibility homogeneous prime, The Axiom of Choice; equality of principal ideals in the polynomial domain makes their generators associates, and its units are the nonzero constants.

An affine plane curve is $V(f)\subseteq {\mathbf{A}}^2$ for a nonconstant square-free $f\in k[x,y]$ An affine algebraic set in affine space, with its components defined by the irreducible factors of $f$ in the same way. On a standard chart $D_{+}(x_i)\cong {\mathbf{A}}^2$ the trace of a projective curve is the zero set of its dehomogenisation, which is an affine curve when the dehomogenisation is nonconstant and is empty when it is a nonzero constant: $C\cap D_{+}(x_i)=V(f_i)$ for $f_i=F(x_i\mapsto 1,x_j\mapsto x_j\mathrm{/}{x}_i\text{ for }j\mathrm{\ne }i)$ projective hypersurface affine pieces, standard projective opens are affine spaces.

Square-freeness and components. The polynomial ring $k[x_0,x_1,x_2]$ is a unique factorisation domain Every finite-variable polynomial ring over a field is a UFD, with prime irreducibles and principal height-one primes, so $F$ has a factorisation into pairwise nonassociate irreducibles, and each factor of a homogeneous form is homogeneous: the lowest and highest nonzero graded parts of a product are the products of the corresponding parts, nonzero in this domain, so a product supported in one degree forces each factor to be supported in one degree Nonnegatively graded rings and modules, homogeneous elements, and twists. Square-freeness of $F$ says that no irreducible factor is repeated: $F=c{\prod }_j{h}_j$ with $c\in k^{\times }$ and the $h_j$ pairwise nonassociate irreducible homogeneous forms. The irreducible components of $C=V(F)$ are the curves $V(h_j)$, one for each distinct factor: the union of the $V(h_j)$ is $C$, each $V(h_j)$ is irreducible and for irreducible homogeneous $h$, the homogeneous vanishing ideal of $V(h)$ is $(h)$ by the radical-ideal correspondence, so up to units and order the list is determined by $F$ and its members are exactly the maximal irreducible closed subsets of $C$ Irreducible topological spaces and irreducible subsets in the subspace topology, projective irreducibility homogeneous prime. Since every $h_j$ occurs exactly once, $C$ is reduced and no component carries an invisible multiplicity in its defining form.

• Nonemptiness of $C$. A nonconstant homogeneous $F$ has a nontrivial zero. First, $k$ is infinite: if $k=\{a_1,\dots ,a_q\}$ were finite, then $p(T)={\prod }_{i=1}^q(T-a_i)+1$ is nonconstant of degree $q\ge 1$ and $p(a)=1\mathrm{\ne }0$ at every $a\in k$, contradicting algebraic closure An algebraically closed field: every nonconstant polynomial has a root in the field, Evaluation and roots of a polynomial in a commutative target ring, Over an integral domain, degrees add under multiplication of nonzero polynomials. Choose a variable, say $x_2$, occurring in $F$ with exponent $r\ge 1$ and write $F={\sum }_{j=0}^r{a}_j(x_0,x_1)x_2^j$ with $a_r\mathrm{\ne }0$; the polynomial $a_r(X,1)\in k[X]$ is then nonzero, so since $k$ is infinite it is not the zero function: a nonzero polynomial of degree $n$ has at most $n$ roots, so it cannot vanish at every element of the infinite field $k$ A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots; hence there is $u\in k$ with $a_r(u,1)\mathrm{\ne }0$. The polynomial $F(u,1,T)\in k[T]$ has degree $r\ge 1$, hence has a root $c$ An algebraically closed field: every nonconstant polynomial has a root in the field, and $[u:1:c]$ is a nonzero point of $V(F)$. For a plane projective curve we use $r\ge 1$, so the argument applies; note $\mathrm{deg}F\ge 1$ throughout.
• Infinitude. Every such curve has an affine chart with a nonconstant equation $f(x,y)$. If $f$ has positive degree in $y$, its top coefficient in $k[x]$ is nonzero and vanishes at only finitely many $a\in k$; for each of the infinitely many remaining $a$, the nonconstant polynomial $f(a,y)$ has a root. If $f$ depends only on $x$, any root gives an entire affine line of zeros. Thus every plane projective curve, including each component, has infinitely many points.
• Scaling and the square-free convention. $V(\lambda F)=V(F)$ for every $\lambda \mathrm{\ne }0$, so the curve does not see the scalar, and writing $F$ without repeated factors is a genuine normalisation: with the nonreduced form $x_0^2$ the line $V(x_0)$ would otherwise be assigned the degree of a nonreduced equation. The definition therefore fixes square-free forms and records reduced components, exactly as degree projective hypersurface does for hypersurfaces.
• Where choice enters. The naming convention for a plane projective curve, the degree of a fixed defining form, and its factorisation are choice-free. The supplied radical-ideal correspondence uses the Axiom of Choice through the Nullstellensatz input of projective irreducibility homogeneous prime The Axiom of Choice. It is used both to recover the defining form up to scalar from the underlying closed set and to identify the $V(h_j)$ as its irreducible components. Consumers invoking either identification, including the resulting well-definedness of degree from the closed set, inherit AC; calculations with a fixed defining form alone need no choice.